Okay, since you're awesome (^.^), I'll not use shortcuts...
So, if we have
\[\large z = r (\cos \theta + i \sin \theta)\]
In general, we get
\[\large z^p=r^p(\cos \ p\theta + i \sin \ p\theta)\]
I say in general, because in the case of fractional exponents (like taking the cube root, for instance), there's a minor catch...

Et voila!
Your three cube roots.
I'll show you a shortcut for this, but first, let's see what happens if continue this process... namely that we take the cube root, only add 360 degrees more to the angle:

Anyway, a shortcut for this, when getting the nth roots of a complex number in polar form, you just apply the rule directly, and only once...
\[\huge 8^{\frac{1}{3}}(\cos \frac{216}{3}^o + i \sin \frac{216}{3}^o)\]
\[\large =2(\cos 72^o + i \sin 72^o)\]
And just keep adding
\[\frac{360}{n}\]
degrees to the angle until you have n distinct answers. In this case, n = 3, so keep addint 360/3 = 120 to the angle until you have three cube roots...
The angles are
72
72 + 120 = 192
72 + 120 + 120 = 312